Question

The integrating factor of differential equation `(1 - y)^2  (dx)/(dy) + yx = ay(-1 < y < 1)`

विकल्प

• `1/(y^2 - 1)`

• `1/sqrt(y^2 - 1)`

• `1/(1 - y^2)`

• `1/sqrt(1 - y^2)`

Answer 1

`1/sqrt(1 - y^2)`

Explanation:

`(1 - y^2)  (dx)/(dy) + yx = 2y` or `(dx)/(dy) + y/(1 - y^2) x = - y/sqrt(1 - y^2)`

∴ `int  y/(1 - y^2)  dy = - 1/2 int (-2y)/(1 - y^2) dy = - 1/2 log (1 - y^2)`

= `log(1 - y)^(1/2) = log(1/sqrt(1 - y^2))`

I.F. = `e^(int  pdx) = e^(log  1/sqrt(1 - y^2)) = 1/sqrt(1 - y^2)`